Show that a uniform limit of bounded functions is bounded.
Show that a uniform limit of continuous function is continuous.
- I.e. if \(f_n\to f\) uniformly with each \(f_n\) continuous then \(f\) is continuous.
Show that
\(f_n: [a, b]\to \RR\) are continuously differentiable with derivatives \(f_n'\)
The sequence of derivatives \(f_n'\) converges uniformly to some function \(g\)
There exists at least one point \(x_0\) such that \(\lim_n f_n(x_0)\) exists,
Then \(f_n \to f\) uniformly to some differentiable \(f\), and \(f' = g\).
Prove that uniform convergence implies pointwise convergence implies a.e. convergence, but none of the implications may be reversed.
Show that \(\sum {x^n \over n!}\) converges uniformly on any compact subset of \(\RR\).
Solution
1. If \(f_n \to f\) uniformly and each \(f_n\) is bounded, then \(f\) is bounded.
Choose \(N\) with \(\|f_N - f\|_\infty < 1\); then \(\|f\|_\infty \le \|f_N\|_\infty + 1\). See E-TZFC7.
2. If \(f_n \to f\) uniformly and each \(f_n\) is continuous, then \(f\) is continuous.
Fix \(x\) and \(\eps > 0\). Choose \(n\) with \(\|f_n - f\|_\infty < \eps/3\), and \(\delta > 0\) with \(|f_n(x) - f_n(y)| < \eps/3\) for \(|x-y| < \delta\). For such \(y\), \(|f(x) - f(y)| \le |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f_n(y) - f(y)| < \eps\). See E-TZFC7.
3. If \(f_n \in C^1[a,b]\), \(f_n' \to g\) uniformly, and \(f_n(x_0)\) converges for some \(x_0\), then \(f_n \to f\) uniformly for a differentiable \(f\) with \(f' = g\).
3.1. \((f_n)\) is uniformly Cauchy, so \(f_n \to f\) uniformly for some continuous \(f\).
By the mean value theorem applied to \(f_m - f_n\), \(|f_m(x) - f_n(x)| \le |f_m(x_0) - f_n(x_0)| + (b-a)\|f_m' - f_n'\|_\infty\) for every \(x\), and both terms tend to \(0\) as \(m, n \to \infty\). A uniformly Cauchy sequence of continuous functions converges uniformly to a continuous function.
3.2. Q.E.D.
\(g\) is continuous as a uniform limit of continuous functions. By the fundamental theorem of calculus \(f_n(t) = f_n(x_0) + \int_{x_0}^t f_n'(s)\,ds\), and uniform convergence passes to the limit: \(f(t) = f(x_0) + \int_{x_0}^t g(s)\,ds\). The fundamental theorem of calculus for continuous \(g\) gives \(f' = g\). See E-7EL3U.
4. Uniform convergence implies pointwise convergence, which implies a.e. convergence, and neither implication reverses.
\(|f_n(x) - f(x)| \le \sup|f_n - f|\) for each \(x\), and a set where convergence fails that is empty is null. On \([0,1]\), \(x^n \to \chi_{\theset{1}}\) pointwise, but \(\sup_{[0,1]}|x^n - \chi_{\theset1}(x)| = 1\) for every \(n\). On \([0,1]\), \(\chi_{[0,1/n)} \to 0\) at every \(x > 0\), so a.e., but \(\chi_{[0,1/n)}(0) = 1\) for every \(n\). See E-UYQED.
5. \(\sum x^n/n!\) converges uniformly on every compact subset of \(\RR\).
A compact set lies in some \([-M, M]\), where \(\left|\frac{x^n}{n!}\right| \le \frac{M^n}{n!}\) and \(\sum_n M^n/n! = e^M < \infty\); the Weierstrass M-test applies.